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svp [43]
3 years ago
12

A ball thrown horizontall at 22.2 m/s from the roof of a building lands 36.0m from the base of the building. How tall is the bui

lding?
Physics
1 answer:
vekshin13 years ago
8 0
Velocity of the ball (v) = 22.2 m/s
Distance from the base of the building where the ball falls (d) = 36.0m
We know that, Velocity (v) = distance (d) / time (t)
From the above equation,
Time (t) = distance (d) / velocity (v)
Substituting the values for distance and velocity in the above equation,
Time = 36.0/22.2 = 1.62 seconds

When the ball is falling down, it accelerates due to gravity
This acceleration (a) = 9.8 m/s²

Now we need to find the height of the building.
The equation of motion for an object with constant acceleration is,
s = ut + 1/2 at²
Where,
s = vertical displacement, which is height of the building in our case
u = initial velocity of the ball = 0

Substituting all the above values in the equation of motion,
s = (0 × 1.62) + (1/2 × 9.8 × 1.62²)
s = 0 + (25.72/2) = 12.86 m = 12.9 m which is the height of the building

Therefore the building is 12.9m tall

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Explanation:

It is given that,

Kinetic energy of the electron, E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.    

The magnetic field in north direction, B_y=19911.5\ nT

The magnetic field in west direction, B_x=-3257.1\ nT

The magnetic field in vertical direction, B_z=48381.8 \ nT

Magnetic field, B=(-3257.1i+19911.5j+48381.8k)\ nT

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

E_k=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E_k}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}

v=5.92\times 10^7 i\ m/s (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

F=q(v\times B)

F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})

Since, i\times j=k\ \\j\times k=i\\k\times i=j

And, i\times i=j\times j=k\times k=0

F=1.6\times 10^{-19}\times [1178 k-2864.20j]

|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}

F=4.95\times 10^{-16}\ N

(b) Let a is the acceleration of the electron. It can be calculated as :

a=\dfrac{F}{m}

a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}

a=5.43\times 10^{14}\ m/s^2

Hence, this is the required solution.

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Answer:

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Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

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